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Old 09-17-2008   #6
ladeewzl
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Default I change my answer!!!!

Ok...so I told this math problem to my husband and he came up with a different (and correct) answer!!

The boy will arrive at the fair with 8 watermelons.

He still travels by the mile and he will make three trips - the first and second trips he eats two watermelons each, but for the third trip (since he doesn't have a return trip to the starting point) he only eats 1 watermelon. So by mile number 1, he has eaten 5 watermelons.

The same thing happens for the trips between miles 2 and 3. So by mile number 3 he has eaten a total of 15 watermelons. Now, he only has to make two trips for each mile (since he only has 30 watermelons left).

Following the same logic as for the first three miles, at miles 4, 5, 6, 7, and 8, he will eat another 15 watermelons (2 for the first trip to the mile and 1 for the last for a total of 3 for each mile). So for those 5 miles, he has eaten another 15 watermelons.

Now he only has 15 watermelons left, 7 miles to go, which leaves him with 8 watermelons at the fair.